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Solution to polynomial of unknown degree 
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#1
Dec213, 04:50 PM

P: 686

Dear!
Is possible to solution a polynomial of kind y = Ax^a + Bx^b ? Thx! 


#2
Dec213, 05:40 PM

P: 84

You have sevens unknowns and one equation.
Could you solve 0 = a + b + c + d + e + f + g + h? 


#3
Dec213, 05:49 PM

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#4
Dec213, 05:56 PM

Sci Advisor
P: 839

Solution to polynomial of unknown degree



#5
Dec313, 12:36 AM

Mentor
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#6
Dec313, 06:02 AM

P: 686

omg, my question is simple!
I would like to know if is possible to isolate the x variable in equation [tex]f(x)=ax^{\alpha}+bx^{\beta}[/tex] 


#7
Dec313, 06:54 AM

HW Helper
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#8
Dec313, 07:14 AM

P: 274

Yes, it's possible, and quite easy. Let us suppose that [itex]\alpha > \beta[/itex]. Let [itex]\zeta = e^{2\pi i/(\alpha  \beta)}[/itex] be a primitive root of unity. Then the polynomial factors as:
[tex]ax^\beta \prod_{k=1}^{\alpha  \beta} (x  \left( \frac{b}{a} \right)^{1/(\alpha  \beta)} \zeta^k)[/tex] 


#9
Dec313, 07:21 AM

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PF Gold
P: 6,733

In mathematics, as in other areas of science, just because a question is 'simple', it does not necessarily follow that the solution will be 'simple'.
For example, consider Fermat's Last Theorem: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem 


#10
Dec313, 07:46 AM

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#11
Dec313, 09:53 AM

P: 686




#12
Dec413, 03:58 AM

P: 1,666

$$f(x,y)=a_1(x)+a_2(x)y+a_3(x)y^2+\cdots+a_n(x)y^n=0$$ with ##a_i(x)## polynomials. In your case we would simply have: $$f(x,y)=xay^{\alpha}by^{\beta}=0$$ Then by NewtonPuiseux's Theorem, we can compute power series representations of the various branches (solutions) of ##y(x)## having the form: $$y_d(x)=\sum_{n=p}^{\infty} c_n\left(x^{1/d}\right)^n$$ with radii of convergences extending at least to the nearest singular point of ##f(x,y)## and often further than that. Do a search for "Newton Polygon" if you're interested in knowing how to compute these "Puiseux" series. 


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